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sritama

  • 4 years ago

a1,a2,a3....a30-each of these 30 sets has 5 elements.b1,b2,....bn-each of these n sets has 3 elements.union of a1,a2...a30=union of b1,b2....bn=S. if each elements of S is in 10 a sets and 9 bsets.then n=? ans=45. show the process

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  1. joemath314159
    • 4 years ago
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    ok, here we go~ So the number of elements in the union of the A sets is: \[5(30)-r_A\]where r is the number of repeats. Likewise the number of elements in the B sets is: \[3n-r_B\] Each element in the union (in S) is repeated 10 times in A, which means if x was the real number of elements in A (not counting repeats) then 9 out of those 10 should be thrown away, or 9x. Likewise on the B side, 8x of those elements should be thrown away. so now we have: \[150-9x = 3n-8x \iff 150-x = 3n \iff 50-\frac{x}{3} = n\] Now, to figure out what x is, we need to use the fact that the union of a group of sets contains every member of each set. if every element in S is repeated 10 times, that means every element in the union of the A's is repeated 10 times. This means that: \[\frac{150}{10} = 15\]is the number of elements in the the A's without repeats counted (same for the Bs as well). So now we have: \[50-\frac{15}{3} = n \iff n = 45\]

  2. sritama
    • 4 years ago
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    i understand.r u a teacher?

  3. joemath314159
    • 4 years ago
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    im an undergrad.

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